Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Construction 2

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Construction for Inscribing Equilateral Triangle inside Square with a Coincident Vertex

Let $\Box ABCD$ be a square.

It is required that $\triangle DGH$ be an equilateral triangle inscribed within $\Box ABCD$ such that vertex $D$ of $\triangle DGH$ coincides with vertex $D$ of $\Box ABCD$.


Construction

Inscribing-equilateral-triangle-inside-square-2.png

By Construction of Equilateral Triangle, let an equilateral triangle $\triangle DAI$ be constructed on $AD$ such that $I$ is inside $\Box ABCD$.

Bisect $\angle ADI$ and bisect it again towards $AD$ to cut $AB$ at $G$.

Construct $H$ on $BC$ such that $DH = DG$.

Then $DGH$ is the required equilateral triangle.


Proof

First a lemma:

Lemma

Let $\Box ABCD$ be a square.

Let $\triangle DGH$ be an isosceles triangle inscribed within $\Box ABCD$ such that the apex $D$ of $\triangle DGH$ coincides with vertex $D$ of $\Box ABCD$.

Inscribing-equilateral-triangle-inside-square-lemma.png

Then:

$\triangle DGH$ is equilateral triangle

if and only if:

$\angle ADG = 15 \degrees \text { or } \angle CDH = 15 \degrees$ (and in fact both are the case).

$\Box$


Because $\triangle DAI$ is equilateral, we have that:

\(\ds \angle ADI\) \(=\) \(\ds 60 \degrees\)
\(\ds \leadsto \ \ \) \(\ds \angle ADG\) \(=\) \(\ds 15 \degrees\) having bisected it twice


The result follows from the lemma.

$\blacksquare$


Sources

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